Question about lowest rung on Harmonic Oscillator I'm reading Griffiths's Introduction to Quantum Mechanics 3rd ed textbook [1]. On p.43, the author explains:

What if I apply the lowering operator repeatedly? Eventually I’m going to reach
a state with energy less than zero, which (according to the general theorem in Problem 2.3) does not exist! At some point the machine must fail. How can that happen?
We know that $a_-ψ$ is a new solution to the Schrödinger equation, but there is no guarantee that it will be normalizable—it might be zero, or its square-integral might be infinite. In practice it is the
former: There occurs a “lowest rung” (call it $ψ_0$) such that
$$a_−ψ_0 = 0 $$

I understood why $a_−ψ_0$ should not be normalized. But why should it be non-normalized like $a_−ψ_0 = 0$? As the author mentioned in the book, the possibility of its square-integral value being infinite may also exist (satisfying with the non-normalizable condition). The author went over this point, and I wonder what happens to the case I mentioned.
Reference
Griffiths, D. J.; Schroeter, D. F. Introduction to Quantum Mechanics 3rd ed; Cambridge University Press, 2018. ISBN 978-1107189638.
 A: $a_-\psi_0$ results in the zero-vector.  Call this vector $\vert\hbox{0 vector}\rangle$.  Then in any computation
$$
\langle \psi_n|\hat T \vert\hbox{0 vector}\rangle =0
$$
for any operator.  In particular the length of $\vert \hbox{0 vector}\rangle$ is $0$ and in this sense it cannot be normalized.
A: I'll be appealing to you physical intuition, so that I don't have dive deep into functional analysis.
Wavefunctions are mathematical objects which are defined in a Hilbert space, which is square integrable. Now, the operators you see in QM are defined on this Hilbert space. Roughly, there are kind of linear functions which map from one Hilbert space to another. Typically, Hamiltonian operators are semi-bounded: there is a lowest energy.
Now, here is the intuition: In the same or previous page, you have the relation $(a_+a_- +\frac{1}{2}\hbar\omega)\psi= E\psi$. If you were to plug $\psi_0$ here and conjecture that $a_-\psi_0$ is infinity, you immediately see that $E$ would also be infinite for the ground state, and thus also for higher states in a quantum harmonic oscillator irrespective of $\omega$. This would also affect the correspondence principle as clearly there exists classical harmonic oscillator, and you won't recover that for any $\omega$ for larger number of states or if $\hbar \rightarrow 0$.
Thus the other possibility of $a_-\psi_0$ being $0$ is reasonable.
Hope that helps.
